A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.
b. What should be its sign, so that all three charges will be in equilibrium?
c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.

Answers

Answer 1

Answer:

Answer:

a) x = ⅔ d, b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

∑ F = 0

-F₁₂ + F₂₃ = 0

F₁₂ = F₂₃

let's replace the values

k Q Q / r₁₂² = k Q 4Q / r₂₃²

Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

r₁₂ = x

r₂₃ = d-x

we substitute

1 / x² = 4 / (d-x) 2

1 / x = 2 / (d-x)

x = 2 (x-d)

x = 2x -2d

3x = 2d

x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c) Q

Answer 2

Answer:

Final answer:

The third charge, -Q, can either be placed at a distance of d/4 to the left of origin or at a distance d/2 to the right of +4Q charge to create an equilibrium among the three charges. Its sign should be negative and its magnitude should be equal to Q.

Explanation:

To create equilibrium among all three charges, our third charge can be placed in two potential positions on the x-axis. One position is a distance of d/4 to the left of the origin (negative x-axis), and the other is a distance of d/2 to the right of the +4Q charge (positive x-axis).

The third charge should be a negative charge, denoted as -Q, in order to balance the positive charges and create equilibrium.

The magnitude of this third charge should be equal to Q, the original charge. The reason is that the forces need to be balanced to create equilibrium and the force is proportional to the charge. Therefore, if -Q is placed at d/4 to the left of the origin or if -Q is placed at d/2 to the right of the +4Q charge, the system will be in equilibrium.

Learn more about Electric Charges and Fields here:

brainly.com/question/18369630

#SPJ3

Related Questions

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?
Which techniques can scientists use to determine the characteristics of Earth's layers? Select the two correct answers.examine the behavior of minerals at extremely low temperaturesstudy how seismic waves travel through different layersdrill deep mines to obtain samples from Earth's mantle and coreconduct experiments about how minerals change under high pressureuse X-rays to obtain a view of Earth's interior layers
Air contains 78.08% nitrogen, 20.095% oxygen, and 0.93% argon. calculate the partial pressure of oxygen if the total pressure of the air sample is 1.7 atm.a.
A 50 kg woman and an 80 kg man stand 12.0 m apart on frictionless ice.(a) How far from the woman is their CM?m(b) If each holds one end of a rope, and the man pulls on the rope so that he moves 1.3 m, how far from the woman will he be now?m(c) How far will the man have moved when he collides with the woman?m
A bullet of mass 0.01 kg moving horizontally strikes a block of wood of mass 1.5 kg which is suspended as a pendulum. The bullet lodges in the wood, and together they swing upwards a distance of 0.40 m. What was the velocity of the bullet just before it struck the wooden block

Bill and Janet are pulling on opposite sides of a table. Bill pulls with a force of450 N to the left, and Janet pulls with a force of 300 N to the right. What is
the net force on the table?
O
A. 450 N to the right
O
B. 450 N to the left
C. 150 N to the left
O
D. 150 N to the right

Answers

Answer:

Explanation:

It is given that,

Force acting in left side, F = 450 N

Force acting in right side, F' = 300 N

Let left side is taken to be negative while right side is taken to be positive. So,

F = -450 N

F' = +300 N

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

$F_(net)=-450\ N+300\ N$

$F_(net)=-150\ N$

So, the net force on the table is 150 N and it is acting to the left side. Hence, the correct option is (c).

Answer:

C. 150 N to the left

Explanation:

If we take right to be positive and left to be negative, then:

∑F = -450 N + 300 N

∑F = -150 N

The net force is 150 N to the left.

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both filaments have the same emissivity, and both bulbs radiate the same power. Find the ratio A1/A2 of the filament areas of the bulbs.

Answers

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$(A_1)/(A_2) =(T_(2)^(4))/(T_(1)^(4))$

substituting the values in the above question we get

$(A_1)/(A_2) =(2100_(2)^(4))/(2700_(1)^(4))$

$(A_1)/(A_2) }$ =0.3659

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 0.90m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.2m from the edge.-How far are the goggles from the edge of the pool?

Answers

Answer:

the googles are 5.3 m from the edge

Explanation:

Given that

depth of pool , d = 3.2 m

Now, let i be the angle of incidence

a laser pointer 0.90 m above the edge of the pool and laser beam enters the water 2.2 m from the edge

⇒tan i = 2.2/0.9

$i=arctan(2.2/.90)$

solving we get

i = 67.8°

Using snell's law ,

n1 ×sin(i) = n2 ×sin(r)

n1= refractive index of 1st medium= 1

n2= refractive index of 2nd medium = 1.33

r= angle of reflection

therefore,

$1* sin(67.8) = 1.33* sin(r)$

r = 44.1°

Now,

distance of googles = 2.2 + d×tan(r)

distance of googles = 2.2 + 3.2×tan(44.1)

distance of googles = 5.3 m

the googles are 5.3 m from the edge

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g= $(372)/(1000)=0.372 Kg$

1 kg=1000g

Maximum acceleration, a= $17.6 m/s^2$

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a= $A\omega^2$

Maximum speed, v= $\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega$

Angular frequency, $\omega=10.06 rad/s$

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=(1.75)/(10.06)=0.17 m$

Hence, the amplitude=0.17 m

c.Spring constant,k= $m\omega^2$

Using the formula

$k=0.372* (10.06)^2$

Hence, the spring constant,k=37.6 N/m

a weight is suspended from the ceiling by a spring (k = 20 ln/in) and is connected to the floor by a dashpot producing viscous damping. The damping force is 10 lb when the velocity of the dashpot plunger is 20 in/sec. The weight and plunger have W = 12 lb. What will be the frequency of the damped vibrations?

Answers

Answer:

The frequency of the damped vibrations is 3.82 Hz.

Explanation:

Given that,

Spring constant = 20 lb/in

Damping force = 10 lb

Velocity = 20 in/sec

Weight = 12 lb

We need to calculate the damping constant

Using formula of damping force

$b* v=F_(d)$

$b=(F_(d))/(v)$

Put the value into the formula

$b =(10)/(20)$

$b=0.5\ lb-sec/in$

$b=0.5*12 =6\ lb-sec/ft$

We need to calculate the frequency

Using formula of angular frequency

$\omga=\sqrt{\omega_(0)^2-((b)/(2m))^2}$

$\omega=\sqrt{(k)/(m)-((b)/(2m))^2}$

Put the value into the formula

$\omega=\sqrt{(20*12*32)/(12)-((6*32)/(2*12))^2}$

$\omega=24\ rad/s$

We need to calculate the frequency of the damped vibrations

Using formula of frequency

$f=(\omega)/(2\pi)$

Put the value into the formula

$f=(24)/(2\pi)$

$f=3.82\ Hz$

Hence, The frequency of the damped vibrations is 3.82 Hz.

A water slide is constructed so that swimmers, starting from rest at the top of the slide, leave the end of the slide traveling horizontally. One person hits the water 5.00 m from the end of the slide in a time of 0.504 s after leaving the slide. Ignore friction and air resistance. Find the height H.

Answers

Answer:

4.93 m

Explanation:

According to the question, the computation of the height is shown below:

But before that first we need to find out the speed which is shown below:

As we know that

$Speed = (Distance)/(Time)$

$Speed = (5)/(0.504)$

= 9.92 m/s

Now

$v^2 - u^2 = 2* g* h$

$9.92^2 = 2* 9.98 * h$

98.4064 = 19.96 × height

So, the height is 4.93 m

We simply applied the above formulas so that the height i.e H could arrive

Final answer:

The height of the water slide is 5.04 meters.

Explanation:

The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.

Other Questions

What can you infer from the fact that metals are good conductors of electricity?

What kind of energy is produce when sun reaches solar panel?

The fundamental force that is responsible for beta decay is the: A) weak force. B) strong force. C) dark energy force. D) gravitational force E) electromagnetic force.

Define the term energy density of a body under strain